A circle with centre (0, 0) passes through the point (5, -12) - Leaving Cert Mathematics - Question 3 - 2010

The standard equation of a circle with center (0, 0) and radius $r$ is given by:

$x^2 + y^2 = r^2.$

Substituting $r = 13$:

$x^2 + y^2 = 13^2 = 169.$

Therefore, the equation of the circle is:

$x^2 + y^2 = 169.$

The line $l$ is given by the equation $x - 4y - 17 = 0$. To find the coordinates of point T, we need to find the intersection of this line with the circle '$x^2 + y^2 = 17$'.

We can isolate $x$ from the line equation:

$x = 4y + 17.$

Now, substitute $x$ in the circle's equation:

$(4y + 17)^2 + y^2 = 17,$

which expands to:

$16y^2 + 136y + 289 + y^2 = 17,$

This simplifies to:

$17y^2 + 136y + 272 = 0.$

Now using the quadratic formula to solve for $y$:

$y = \frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} = \frac{-136 \\pm \\sqrt{136^2 - 4 \times 17 \times 272}}{2 \times 17}.$

Calculating the values will provide two possible coordinates for T.

After finding the coordinates of T as $(x_T, y_T)$, the other end-point can be located by analyzing the midpoint between the center of the circle $(0, 0)$ and point T. If we denote the other point as $(x_O, y_O)$, the midpoint formula gives the relationship:

$\left(\frac{x_T + x_O}{2}, \frac{y_T + y_O}{2}\right) = (0,0).$

From which we can find that:

$x_O = -x_T, \quad y_O = -y_T.$

This results in the coordinates for the other end-point.

The circle has the equation $x^2 + (y - 7)^2 = 100$. From this, we can deduce:

• The center of the circle is at (0, 7).
• The radius can be calculated as $\\sqrt{100} = 10$.

The point (6, h) lies on the circle, which can be confirmed by substituting it into the circle's equation:

$6^2 + (h - 7)^2 = 100.$

This simplifies to:

$36 + (h - 7)^2 = 100,$

leading to:

$(h - 7)^2 = 64.$

Taking the square root gives:

$h - 7 = 8 \\text{ or } h - 7 = -8.$

Thus, the potential values for h are:

$h = 15 \\text{ and } h = -1.$